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Hello StackExchange users. I have discovered a formula which works out any Fibonacci number using the formula to work out the value of any cell in Pascal's triangle. It uses the sigma notation which I don't know whether or not I can remove to condense the formula down but here goes:

$$F(n)=\sum_{z=0}^l \frac{(n-z-1)!}{z!(n-2z-1)!}$$ where $F(n)$ is the Fibonacci sequence

where $l = rnd(\frac{n}{2}) - 1$

where $rnd(x)$ is the function to round odd results of $\frac{n}{2}$ to a whole number

If there is any way to condense this, please post below. Also, if you find any errors, i.e. it doesn't work for a particular number, please post below.

My name is Taylor Golden, thank you for reading and I hope that, if any of you are working with the Fibonacci sequence, that this helps you. Please remember to give credit!

I should also note how I got to this. I was working with Pascal's triangle and discovered the formula to find any value of a cell which is:

$${r \choose t}=\frac{r!}{t!(r-t)!}$$

Using this, on the same website was also how to find the Fibonacci numbers on Pascal's triangle. I found that there was a pattern between the intervals of the cells on the triangle and thus, used this to my advantage to gain the above formula.

Thank you again!

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closed as unclear what you're asking by Hayden, alexqwx, naslundx, Kirill, Behaviour Jul 11 at 0:30

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question.If this question can be reworded to fit the rules in the help center, please edit the question.

    
What is the question? –  Hayden Jul 10 at 21:23
    
@Hayden There is a task and a question; my task is to help me find out if there are any numbers which do not work with this formula and my question is whether or not this can be condensed to a formula which is a little more pleasing to the eye. –  capturographer Jul 10 at 21:25
    
If $n$ is even, then rounding $n/2$ is unnecessary. If $n$ is odd, then it is ambiguous what rounding $n/2$ to "the nearest integer" should mean, since it is equal distance from the two nearest integers that bracket it. –  hardmath Jul 10 at 22:49

1 Answer 1

up vote 7 down vote accepted

So that would be $$ F(n) = \sum_{z=0}^l{n-z-1\choose z},$$ right? As a matter of fact, this result is already apparent in (my view of) google's first rsult page - see the image on the right. A proof using the recursive formulas of binomial coefficients and of Fibonacci numbers is straightforward. Sorry, this is not a new result, but congratulations on (re-)discovereing it!

enter image description here

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Oh, I see. Thank you. I didn't realise this. It only took me an hour through Statistics class - I thought it couldn't have been that hard to find! –  capturographer Jul 10 at 21:36
2  
+1: Very gentle. –  copper.hat Jul 10 at 21:49

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